A body is projected vertically upward with speed `10 m//s` and other at same time with same speed in downward direction from the top of a tower. The magnitude of acceleration of first body w.r.t. second is {take `g = 10 m//s^(2)`}

Options

A. Zero
B. `10 m//s^(2)`
C. `5m//s^(2)`
D. `20m//s^(2)`

Answer

To solve the problem, we need to determine the magnitude of acceleration of the first body (projected upward) with respect to the second body (projected downward).

### Step-by-Step Solution:

1. **Identify the Acceleration of Each Body**:
- The first body (let's call it Body A) is projected vertically upward with an initial speed of \(10 \, \text{m/s}\). However, it will experience a downward acceleration due to gravity, which is \(g = 10 \, \text{m/s}^2\).
- The second body (let's call it Body B) is projected downward with the same initial speed of \(10 \, \text{m/s}\) and will also experience the same downward acceleration due to gravity, \(g = 10 \, \text{m/s}^2\).

2. **Determine the Direction of Accelerations**:
- For Body A (upward motion), the acceleration due to gravity acts downward, which we can denote as \(a_A = g = 10 \, \text{m/s}^2\).
- For Body B (downward motion), the acceleration due to gravity also acts downward, so we denote \(a_B = g = 10 \, \text{m/s}^2\).

3. **Calculate the Relative Acceleration**:
- The relative acceleration of Body A with respect to Body B is given by the formula:
\[ a_{AB} = a_A - a_B \]
- Substituting the values we have:
\[ a_{AB} = 10 \, \text{m/s}^2 - 10 \, \text{m/s}^2 = 0 \, \text{m/s}^2 \]

4. **Conclusion**:
- The magnitude of the acceleration of the first body with respect to the second body is \(0 \, \text{m/s}^2\).

### Final Answer:
The magnitude of acceleration of the first body with respect to the second body is \(0 \, \text{m/s}^2\).